Question 144837

Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{x^2+4=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{x^2+0*x+4=0}}}  notice {{{a=1}}}, {{{b=0}}}, and {{{c=4}}})





{{{x = (0 +- sqrt( (0)^2-4*1*4 ))/(2*1)}}} Plug in a=1, b=0, and c=4




{{{x = (0 +- sqrt( 0-4*1*4 ))/(2*1)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+-16 ))/(2*1)}}} Multiply {{{-4*4*1}}} to get {{{-16}}}




{{{x = (0 +- sqrt( -16 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (0 +- 4*i)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (0 +- 4*i)/(2)}}} Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


{{{x=0 + 2*i}}} or {{{x=0 - 2*i}}}


Notice if we graph the quadratic {{{y=x^2+4}}}, we get


{{{ graph( 500, 500, -15, 15, -11, 19, x^2+4) }}} graph of {{{y=x^2+4}}}


We can see that there are no real roots